Newton's method and Mean Value theorem

[Lily@pie]

Homework Statement

let x₀, x₁,... be the approximations of pi from the Newton's Method. Use Mean Value theorem to show that
|pi-x_j+1|=|tan²c_j||pi-x_j|
for some c_j between x_j and pi

Homework Equations

pi is defined as smallest positive number r when sin r =0

The Attempt at a Solution

I have tried to let f(x) = sin x and the bound to be [x,pi].
By using the mean value theorem, there exist some c such that
f'(c) = (sin pi - sin x) / (pi - x)
cos c = - sin x / (pi-x)
But I couldn't get the form, especially the |tan²c_j| part.

I have also tried to let f(x) = tan x - x in [x,pi]
so, f'(c)=sec²c + 1 = (tan pi + pi - tan x - x)/(pi-x)
tan² c = (tan pi - tan x + pi - x)/(pi-x)
But tan pi is undefined...

Am I on the correct path?

 

[snipez90]

I have also tried to let f(x) = tan x - x in [x,pi]
so, f'(c)=sec²c + 1 = (tan pi + pi - tan x - x)/(pi-x)
tan² c = (tan pi - tan x + pi - x)/(pi-x)
But tan pi is undefined...

Am I on the correct path?

This second route is a good path. Note tan(pi) = sin(pi)/cos(pi) = 0/(-1) = 0. Then you're done, assuming you know what Newton's method is.

This was actually a decent problem.

 

[Lily@pie]

I've used f(x) = tan x - x

and manage to find
tan²c_j= (-pi + x_j - tan x_j)/(pi - x_j)

From intermediate theorem,
x_j+1 = x_j - (tan x_j - x_j)/tan²x_j

Hence, I get
tan²c_j
= (-pi + x_j+1 - tan x_j + (tan x_j - x_j)/tan²x_j))/ (pi-x_j)

I have tried all sorts of method to evaluate this but still couldn't get the form (pi-x_j+1)/(pi-x_j)...

I have tried opening the tan x, but still stuck...

Did I do it correctly? Do you mind providing any hints?

 

[snipez90]

You shouldn't go back to the intermediate value theorem. The problem statement tells you that we're approximating pi with Newton's method. The fact that pi is defined as smallest positive number r when sin r = 0 makes it clear how the sequence that defines Newton's method is generated here. We are approximating a root of sin(x) basically. Now write out the sequence that defines Newton's method.

 

[Lily@pie]

Oh! Thank you so much!